More probability. (Simple?) (1 Viewer)

[Hotdog1]

actually i'm interested in this too, and i don't really have the time right now to go through three threads and link everything together, but i did anyway and i still dont see the connection to your proposed solution. Maybe im stupid, but can't you just roughly explain your reasoning. Since you found the time to write such a long reply, made up 2 related questions, replied to multiples of questions, and since you posted the solution on the last page, obviously found the TIME to work it out, so why not just explain it here?

maybe we are not as smart as you are and can work everything out in 1 line...

 

[maniacguy]

Consider A A B C D
There are (5P5/2 - 4P4) ways to arrange these without the two A's touching.

Now we seek to put in another three letters (B,C,D) and to make sure of the order they are in. This should give the other factor, as I understand Richard's solution.

 

[Hotdog1]

Oh alrightee, I think I understand, I think. hehe

 

[Bannanafish]

Originally posted by maniacguy
Consider A A B C D
There are (5P5/2 - 4P4) ways to arrange these without the two A's touching.

Now we seek to put in another three letters (B,C,D) and to make sure of the order they are in. This should give the other factor, as I understand Richard's solution.

that's what i was thinking, but the thing is there are more factors to that